Which of the following compounds will be colored in the solid state?

The correct answer is Option (2) → $CuF_2$

Core Concept:

Color in transition metal compounds arises due to $d-d$ electronic transitions.

• If the metal ion has partially filled d-orbitals $\rightarrow$ compound is colored.
• If the metal ion has completely filled or empty d-orbitals $\rightarrow$ compound is colorless.

Option-wise Detailed Explanation:

Option 1: $Ag_2SO_4$

Silver in $Ag_2SO_4$ is in $Ag^+$ state.

Electronic configuration of $Ag = [Kr]4d^{10}5s^1$.

$Ag^+$ loses $5s$ electron $\rightarrow Ag^+ = 4d^{10}$.

• Completely filled d-orbitals.
• No $d-d$ transition possible.
• Hence, no absorption in visible region.

Compound remains colorless in solid state.

Option 2: $CuF_2$

Copper in $CuF_2$ is in $Cu^{2+}$ state.

Electronic configuration of $Cu = [Ar]3d^{10}4s^1$.

$Cu$ loses $2$ electrons ($1$ from $s$ and $1$ from $d$ orbital).

$\rightarrow Cu^{2+} = 3d^9$.

• Partially filled d-orbitals.
• $d-d$ transitions possible.
• Absorbs visible light.
• Compound appears colored (bluish/greenish).

Option 3: $ZnF_2$

Zinc in $ZnF_2$ is in $Zn^{2+}$ state.

Electronic configuration of $Zn = [Ar]3d^{10}4s^2$.

$Zn$ loses $2$ electrons from $s$ orbital $\rightarrow Zn^{2+} = 3d^{10}$.

• Completely filled d-orbitals.
• No $d-d$ transition possible.
• Hence, no visible light absorption.

Compound remains colorless.

Option 4: $Cu_2Cl_2$

Copper in $Cu_2Cl_2$ is in $Cu^+$ state.

$Cu$ loses $1$ electron $\rightarrow Cu^+ = 3d^{10}$.

• Fully filled d-subshell.
• No $d-d$ transitions.
• Hence, no visible color.

Compound is colorless.